Calculus

Application of Derivatives

Easy

applied

The total cost $C(x)$ in Rupees associated with the production of $x$ units of an item is given by $C(x) = 0.007x^3 + 26x^2 + 15x + 400$ . The marginal cost when 10 items are produced is:

₹ 537.1

₹ 441.15

₹ 1575

₹ 875.25

✅ Correct Option: 1

The total cost function is $C(x) = 0.007x^3 + 26x^2 + 15x + 400$ .

Marginal cost is the derivative of the total cost function.

$C'(x) = \dfrac{d}{dx}(0.007x^3 + 26x^2 + 15x + 400)$

$C'(x) = 0.007 \times 3x^2 + 26 \times 2x + 15 + 0$

$C'(x) = 0.021x^2 + 52x + 15$

To find the marginal cost when 10 items are produced, substitute $x = 10$ into $C'(x)$ :

$C'(10) = 0.021(10)^2 + 52(10) + 15$

$C'(10) = 0.021(100) + 520 + 15$

$C'(10) = 2.1 + 520 + 15$

$C'(10) = 537.1$

Therefore, the marginal cost when 10 items are produced is ₹537.1.

2022: 25 May Shift 1

The equation of the tangent, to the curve $y = x^2 - 2x - 3$ which is perpendicular to the line $x + 2y + 3 = 0$ , is

2025: 27 May Shift 1

For $x \in \mathbb{R}$ , if $f(x) = -(x-1)^2 + 2$ , then

(A) $f$ is an increasing function on $(-\infty, 1]$

(B) $f$ has no critical points

(D) $f$ has a minimum value at $x = 1$

Choose the correct answer from the options given below:

2025: 19 May Shift 1

The minimum value of the function $f(x) = x^3 + (10-x)^3$ occurs at:

2022: 23 Aug Shift 1

Distance between Sumit and Amit in terms of $x$ is :

2025: 13 May Shift 1

The function $f(x) = x + \frac{a^2}{x}$ , $a > 0$ , $x \neq 0$ has a local maxima at

2023: 30 May Shift 3

If the function $f(x) = x^4 - 62x^2 + ax + 9$ attains its local maximum value at $x = 1$ , then a is equal to :

The total cost C(x)C(x)C(x) in Rupees associated with the production of xxx units of an item is given by C(x)=0.007x3+26x2+15x+400C(x) = 0.007x^3 + 26x^2 + 15x + 400C(x)=0.007x3+26x2+15x+400. The marginal cost when 10 items are produced is: